Question

In a hurdle race, a runner has probability $p$ of jumping over a specific hurdle. Given that in $5$ trials, the rummer succeeded $3$ times, the conditional probability that the runner had succeeded in the first trial is

• A. $\frac{3}{5}$
• B. $\frac{2}{5}$
• C. $\frac{1}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A $\frac{3}{5}$

Let $A$ denote the event that the runner succeeds exactly $3$ times out of five and $B$ denote the event that the number succeeds on the first trial.
$P\left(B|A\right)=\frac{P(B\cap A)}{P\left(A\right)}$
But $P(B\cap A)=P$ (succeeding in the first trial and exactly once in two other trials)
$=p{(}^4{C}_2{p}^2(1-p{)}^2)=6p^3(1-p{)}^2$
and $P\left(A\right){=}^5{C}_3{p}^3(1-p{)}^2=10p^3(1-p{)}^2$
Thus, $P\left(B|A\right)=\frac{6p^3(1-p{)}^2}{10p^3(1-p{)}^2}=\frac{3}{5}$.